1981 AHSME Problems/Problem 18

Problem:

The number of real solutions to the equation $\dfrac{x}{100}=\sin x$ is

$\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65$

Solution:

The answer to this problem is the number of intersections between the graph of $f(x) = \sin x$ and $f(x) = \frac{1}{100}x.$ We can do the right side of the coordinate plane first. Each cycle of the sine wave, consisting of 2π, will have 2 intersections (From the positive part of the sine wave) The line $f(x) = \frac{1}{100}x$ will consist of 16 cycles plus a little bit extra for $x$ from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is $2 \cdot 16 = 32.$ Because of symmetry, we can multiply by two to account for the left side, and subtract one because of the origin. So the answer is $32 \cdot 2 - 1 = \textbf{(C)}\ 63.$

~Eric X

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1981 AHSME Problems/Problem 18

Problem:

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